Pullback of the Frobenius map and function fields extension We know that for two nonsingular projective algebraic curves defined over a field $\mathbf{k}$, there is a bijection between nonconstant morphisms $h: C \to C'$ and function fields injections $h^* : \mathbf{k}(C') \to \mathbf{k}(C)$, where $h^*G = G \circ h$.
Also, if $C$ is defined over $\overline{\mathbf{F}}_p$, the Frobenius map on $C$ is 
$$ \sigma_p : C \to C^{\sigma_p}, \quad [x_0:\dots:x_n] \mapsto [x_0^p:\dots:x_n^p]. $$
Now, the example I'm looking at is $C = \mathbf{P}^1(\overline{\mathbf{F}}_p)$, with $\sigma_p(t) = t^p$ on the affine part. Diamond and Shurman say that this gives the extension of function fields $\mathbf{F}_p(t)/\mathbf{F}_p(s)$, where $s = t^p$ and this is what I don't understand. Why is the function field $\mathbf{K}$ of $C$ not $\overline{\mathbf{F}}_p(t)$, and how does one get $\mathbf{F}_p(s)$ for the function field of $C^{\sigma_p}$? Do we need to compute the function field explicitly or the preimage of the pullback map $\sigma_p^*$ instead? I would guess the second but it's not very clear to me how this is done exactly.
 A: Let me start by saying that I understand your confusion and I hope this answer can clear it up. However, it is somewhat of a fine point and if I fail, please point that out.
You have a morphism $\sigma_p: \Bbb P^1 \to \Bbb P^1$. Let us denote the function field of the right side by $\overline{\Bbb F}_p(t)$ and the function field of the one on the left side (domain of $\sigma_p$) by $\overline{\Bbb F}_p(s)$. They are in fact the same field when considered independently, i.e. $s$ and $t$ play the same role for each of them respectively. However, the different variables will help us keep track of what is happening when we see them in relation to one another.
Take the function $t\in\overline{\Bbb F}_p(t)$, a rational function on the image of $\sigma_p$. We then compute its image under the map $\sigma_p^\ast$, which is a function on the domain of $\sigma_p$:
$$
\sigma_p^\ast(t)=t\circ \sigma_p=s^p
$$
This is true because for any point $x\in\Bbb A^1\subseteq\Bbb P^1$, we have $t(\sigma_p(x))=t(x^p)=x^p=s(x)^p$. Now this means that the map $\sigma_p^\ast:\overline{\Bbb F}_p(t)\to\overline{\Bbb F}_p(s)$ which defines your field extension is given by mapping $t$ to $s^p$, because there only is one field homomorphism over $\overline{\Bbb F}_p$ that does so: This follows from the fact that there is only one $\overline{\Bbb F}_p$-algebra homomorphism $\overline{\Bbb F}_p[t]\to\overline{\Bbb F}_p(s)$ with $t\mapsto s^p$, the universal property of the polynomial ring.
So maybe a better way for you to understand this homomorphism is to accept $s$ as the "smallest" variable in all the universe and then define $t:=s^p$. Then, the next important realization is the fact that $\overline{\Bbb F}_p(t)$ is still a function field in one variable, even if it is "just" $\overline{\Bbb F}_p(s^p)$.
For a more concrete and different example, the fields $\Bbb Q(\pi)$ and $\Bbb Q(\pi^8)$ are both isomorphic to $\Bbb Q(t)$, where $t$ is a formal variable: That's because both $\pi$ and $\pi^8$ are transcendental. So is $t$, and so is $t^{23453}$. There are a lot of things that behave like an independent variable over $\Bbb Q$, or over $\overline{\Bbb F}_p$ for that matter. 
